The sum of 10 terms of the series frac{3}{1^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The $n^{th}$ term of the series is given by $T_n = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$.For $n=1, 2, \ldots, 10$,

Vedclass · Accepted Answer

$\frac{120}{121}$

Vedclass · Answer

(B) The $n^{th}$ term of the series is given by $T_n = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$.For $n=1, 2, \ldots, 10$,the sum $S_{10}$ is:$S_{10} = \sum_{n=1}^{10} \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$.This is a telescoping series:$S_{10} = \left( \frac{1}{1^2} - \frac{1}{2^2} \right) + \left( \frac{1}{2^2} - \frac{1}{3^2} \right) + \ldots + \left( \frac{1}{10^2} - \frac{1}{11^2} \right)$.All intermediate terms cancel out:$S_{10} = 1 - \frac{1}{11^2} = 1 - \frac{1}{121} = \frac{120}{121}$.

The sum of $10$ terms of the series $\frac{3}{1^{2} \times 2^{2}}+\frac{5}{2^{2} \times 3^{2}}+\frac{7}{3^{2} \times 4^{2}}+\ldots$ is :

Similar Questions

$\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \ldots + \frac{1}{(3n-1)(3n+2)}$ is equal to

The value of $\sum_{k=1}^{13} \frac{1}{\sin \left(\frac{\pi}{4}+\frac{(k-1) \pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$ is equal to

The sum of the infinite series $\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \frac{4}{5!} + \dots$ is

$\lim _{n \rightarrow \infty} \left( \frac{1}{3 \cdot 7} + \frac{1}{7 \cdot 11} + \frac{1}{11 \cdot 15} + \ldots + n \text{ terms} \right) =$

Find the sum of the $n$ terms of the series $\frac{3}{1^2} + \frac{5}{1^2 + 2^2} + \frac{7}{1^2 + 2^2 + 3^2} + ...$

Vedclass Test Series

Exam Paper Generator

Online Exam Module